1. Simplify $\sqrt[3]{80}$. - Step 1: Factor 80 into prime factors: $80 = 2^4 \times 5$. - Step 2: Use the property $\sqrt[3]{a^3 \times b} = a \sqrt[3]{b}$. - Step 3: Extract $2^3$ from $2^4$: $\sqrt[3]{2^4 \times 5} = \sqrt[3]{2^3 \times 2 \times 5} = 2 \sqrt[3]{10}$. Final answer: $2 \sqrt[3]{10}$. 2. Simplify $\sqrt[4]{162 n^7 p^{11}}$. - Step 1: Factor 162: $162 = 2 \times 3^4$. - Step 2: Express powers inside the fourth root: $\sqrt[4]{2 \times 3^4 \times n^7 \times p^{11}}$. - Step 3: Extract perfect fourth powers: $3^4 = (3)^4$, $n^7 = n^{4+3} = n^4 \times n^3$, $p^{11} = p^{8+3} = p^8 \times p^3$. - Step 4: Apply root extraction: $\sqrt[4]{3^4} = 3$, $\sqrt[4]{n^4} = n$, $\sqrt[4]{p^8} = p^2$. - Step 5: Remaining inside root: $\sqrt[4]{2 \times n^3 \times p^3}$. Final answer: $3 n p^2 \sqrt[4]{2 n^3 p^3}$. 3. Simplify $\sqrt[3]{\frac{3}{8}}$. - Step 1: Use property $\sqrt[3]{\frac{a}{b}} = \frac{\sqrt[3]{a}}{\sqrt[3]{b}}$. - Step 2: $\sqrt[3]{3} / \sqrt[3]{8} = \frac{\sqrt[3]{3}}{2}$ since $\sqrt[3]{8} = 2$. Final answer: $\frac{\sqrt[3]{3}}{2}$. 4. Simplify $\sqrt[3]{\frac{2a}{25c^2}}$. - Step 1: Apply cube root to numerator and denominator separately: $\frac{\sqrt[3]{2a}}{\sqrt[3]{25 c^2}}$. - Step 2: Factor denominator: $25 = 5^2$. - Step 3: No perfect cubes to extract, so expression remains as is. Final answer: $\frac{\sqrt[3]{2a}}{5 \sqrt[3]{c^2}}$. 5. Simplify $\sqrt[4]{81 m^{10} p^6}$. - Step 1: Factor 81: $81 = 3^4$. - Step 2: Express powers: $m^{10} = m^{8+2}$, $p^6 = p^{4+2}$. - Step 3: Extract perfect fourth powers: $\sqrt[4]{3^4} = 3$, $\sqrt[4]{m^8} = m^2$, $\sqrt[4]{p^4} = p$. - Step 4: Remaining inside root: $\sqrt[4]{m^2 p^2}$. Final answer: $3 m^2 p \sqrt[4]{m^2 p^2}$. 6. Simplify $\frac{x}{\sqrt{x-1}}$. - Step 1: Rationalize denominator by multiplying numerator and denominator by $\sqrt{x-1}$. - Step 2: $\frac{x}{\sqrt{x-1}} \times \frac{\sqrt{x-1}}{\sqrt{x-1}} = \frac{x \sqrt{x-1}}{x-1}$. Final answer: $\frac{x \sqrt{x-1}}{x-1}$. 7. Simplify $\frac{1}{\sqrt{3} - 4}$. - Step 1: Rationalize denominator by multiplying numerator and denominator by conjugate $\sqrt{3} + 4$. - Step 2: $\frac{1}{\sqrt{3} - 4} \times \frac{\sqrt{3} + 4}{\sqrt{3} + 4} = \frac{\sqrt{3} + 4}{(\sqrt{3})^2 - 4^2} = \frac{\sqrt{3} + 4}{3 - 16} = \frac{\sqrt{3} + 4}{-13}$. Final answer: $-\frac{\sqrt{3} + 4}{13}$. 8. Simplify $\frac{\sqrt{3}}{\sqrt{3} - \sqrt{2}}$. - Step 1: Rationalize denominator by multiplying numerator and denominator by conjugate $\sqrt{3} + \sqrt{2}$. - Step 2: $\frac{\sqrt{3}}{\sqrt{3} - \sqrt{2}} \times \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} = \frac{\sqrt{3}(\sqrt{3} + \sqrt{2})}{3 - 2} = \sqrt{3}(\sqrt{3} + \sqrt{2})$. - Step 3: Simplify numerator: $\sqrt{3} \times \sqrt{3} = 3$, $\sqrt{3} \times \sqrt{2} = \sqrt{6}$. Final answer: $3 + \sqrt{6}$.